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Home > Operation Research calculators > Graphical method example
1. Graphical method example ( Enter your problem )
( Enter your problem )
  1. Elements of LPP and Definition
  2. Algorithm & Example-1
  3. Maximization Example-2
  4. Maximization Example-3
  5. Minimization Example-4
  6. Minimization Example-5
  7. Mixed constraints Example-6
  8. Mixed constraints Example-7
  9. Multiple optimal solution example
  10. Unbounded solution example
  11. Infeasible solution example
 		Other related methods
  1. Formulate linear programming model
  2. Graphical method
  3. Simplex method (BigM method)
  4. Two-Phase method
  5. Primal to dual conversion
  6. Dual simplex method
  7. Integer simplex method
  8. Branch and Bound method
  9. 0-1 Integer programming problem
  10. Revised Simplex method
7. Mixed constraints Example-6
(Previous example)		9. Multiple optimal solution example
(Next example)

8. Mixed constraints Example-7


Find solution using graphical method
MIN z = 20x1 + 10x2
subject to
x1 + 2x2 <= 40
3x1 + x2 >= 30
4x1 + 3x2 >= 60
and x1,x2 >= 0

Solution:
Problem is
MIN `z_x``=````20``x_1`` + ``10``x_2`
subject to
`````x_1`` + ``2``x_2`≤`40`
```3``x_1`` + ````x_2`≥`30`
```4``x_1`` + ``3``x_2`≥`60`
and `x_1,x_2 >= 0; `



Hint to draw constraints

1. To draw constraint `color{red}{x_1+2x_2<=40 ->(1)}`

Treat it as `color{red}{x_1+2x_2=40}`

When `x_1=0` then `x_2=?`

`=>(0)+2x_2=40`

`=>2x_2=40`

`=>x_2=(40)/(2)=20`

When `x_2=0` then `x_1=?`

`=>x_1+2(0)=40`

`=>x_1=40`

`x_1`040
`x_2`200



2. To draw constraint `color{green}{3x_1+x_2>=30 ->(2)}`

Treat it as `color{green}{3x_1+x_2=30}`

When `x_1=0` then `x_2=?`

`=>3(0)+x_2=30`

`=>x_2=30`

When `x_2=0` then `x_1=?`

`=>3x_1+(0)=30`

`=>3x_1=30`

`=>x_1=(30)/(3)=10`

`x_1`010
`x_2`300



3. To draw constraint `color{blue}{4x_1+3x_2>=60 ->(3)}`

Treat it as `color{blue}{4x_1+3x_2=60}`

When `x_1=0` then `x_2=?`

`=>4(0)+3x_2=60`

`=>3x_2=60`

`=>x_2=(60)/(3)=20`

When `x_2=0` then `x_1=?`

`=>4x_1+3(0)=60`

`=>4x_1=60`

`=>x_1=(60)/(4)=15`

`x_1`015
`x_2`200








The value of the objective function at each of these extreme points is as follows:
Extreme Point
Coordinates
(`x_1`,`x_2`)		Lines through Extreme PointObjective function value
`z=20x_1 + 10x_2`
`color{red}{A(4,18)}``color{red}{1->x_1+2x_2<=40}`
`color{green}{2->3x_1+x_2>=30}`		`20(4)+10(18)=260`
`color{green}{B(6,12)}``color{green}{2->3x_1+x_2>=30}`
`color{blue}{3->4x_1+3x_2>=60}`		`20(6)+10(12)=240`
`color{blue}{C(15,0)}``color{blue}{3->4x_1+3x_2>=60}`
`color{black}{5->x_2>=0}`		`20(15)+10(0)=300`
`color{brown}{D(40,0)}``color{red}{1->x_1+2x_2<=40}`
`color{black}{5->x_2>=0}`		`20(40)+10(0)=800`


The miniimum value of the objective function `z=240` occurs at the extreme point `(6,12)`.

Hence, the optimal solution to the given LP problem is : `x_1=6, x_2=12` and min `z=240`.

This material is intended as a summary. Use your textbook for detail explanation.
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7. Mixed constraints Example-6
(Previous example)		9. Multiple optimal solution example
(Next example)


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